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Question-185231




Question Number 185231 by Rupesh123 last updated on 18/Jan/23
Answered by Kalebwizeman last updated on 19/Jan/23
70 degrees
$$\mathrm{70}\:{degrees} \\ $$
Answered by Kalebwizeman last updated on 19/Jan/23
draw a line joining OQ  angle POQ is 40 (same segment)  and triangle POQ is isosceles  so the required angle just one of the base angles
$${draw}\:{a}\:{line}\:{joining}\:{OQ} \\ $$$${angle}\:{POQ}\:{is}\:\mathrm{40}\:\left({same}\:{segment}\right) \\ $$$${and}\:{triangle}\:{POQ}\:{is}\:{isosceles} \\ $$$${so}\:{the}\:{required}\:{angle}\:{just}\:{one}\:{of}\:{the}\:{base}\:{angles} \\ $$
Commented by HeferH last updated on 19/Jan/23
but how POQ is 40° ?
$${but}\:{how}\:{POQ}\:{is}\:\mathrm{40}°\:? \\ $$
Answered by mr W last updated on 20/Jan/23
Commented by mr W last updated on 20/Jan/23
since 2×70+40=180°  ⇒OPQR is cyclic.  ∠OPQ=70°
$${since}\:\mathrm{2}×\mathrm{70}+\mathrm{40}=\mathrm{180}° \\ $$$$\Rightarrow{OPQR}\:{is}\:{cyclic}. \\ $$$$\angle{OPQ}=\mathrm{70}° \\ $$

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